Classify the singularity $z_0=0$ of $f(z)=\dfrac{z}{z^2+\sin^3z}$.
I found that $$ \lim_{z \to 0} zf(z) =\lim_{z \to 0} \dfrac{z^2}{z^2+\sin^3z}=1, $$ so I have a pole of order $1$. Am I right? If yes, then am I ok to tell that the term of order $-1$ is $1$ and the term of order $-2$ is $0$?
Yes, you're right: the pole has order $1$. In order to find the residue, you can compute $a$ such that $$ \lim_{z\to0}\left(\frac{z}{z^2+\sin^3z}-\frac{a}{z}\right) $$ is finite. The expression becomes $$ \frac{z^2-az^2-a\sin^3z}{z(z^2+\sin^3z)} $$ If $a=1$, then the limit is $-1$; therefore the residue is $1$.